Question 188506
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There are two (immediately obvious) ways to do this.


Let <i>n</i> be the row number.


Then if row 1 has 20, row 2 has 20 + 5(1), and row 3 has 20 + 5 + 5 = 20 + 5(2), you can say that row n has 20 plus 5 times (the row number minus 1).


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  S_n = 20 + 5(n - 1)]


Let's check for row 20:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  S_{20} = 20 + 5(20 - 1) = 20 + 95 = 115]


The other way to do it is to consider an imaginary row 0 that has 5 less seats than the first row, namely 15.  Then the first row has 15 plus 5(1) = 20, the second row has 15 + 5(2) = 25, and the nth row has 15 + 5n


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  S_n = 15 + 5n]


Again, let's try row 20:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  S_n = 15 + 5(20) = 15 + 100 = 115]


Same correct result.




John
*[tex \LARGE e^{i\pi} + 1 = 0]
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